# Restricting a block operator matrix to a subspace

In the answer to the question What is meant by a matrix being strictly positive definite on its range? the use who answered provides the following procedure for restricting a given scalar matrix to a particular subspace.

I recap the answer provided there here: suppose $$M\in\mathbb{R}^{n\times n}$$ is a scalar matrix with with $$k eigenvalues at $$0$$. Suppose, without loss of generality, that $$M$$ is of the form $$M=\begin{bmatrix}0_{k\times k} & 0_{k\times(n-k)}\\0_{(n-k)\times k} & M_p\end{bmatrix}$$ with the block $$M_p$$ positive definite. Then, the range of $$M$$ is spanned by the vector $$V$$ where $$V=\begin{bmatrix}0_{k\times (n-k)}\\ I_{n-k} \end{bmatrix}.$$ The restriction of the matrix $$M$$ to the subspace $$V$$ is then given as $$M|_V = V^TMV=M_p.$$ If $$M$$ were instead a block operator matrix, with entries as bounded linear operators on Hilbert spaces, would the same procedure for restricting the block operator to a subspace work? More precisely, what I mean in this setting is that $$M\in L(\mathcal{H})$$ is of the form $$M=\begin{bmatrix}0_{k\times k} & 0_{k\times(n-k)}\\0_{(n-k)\times k} & M_p\end{bmatrix},$$ where $$\mathcal{H}=H_1\oplus H_2\oplus \dots \oplus H_n$$ is the direct sum of the Hilbert spaces $$\left(H_{i}\right)_{i=1}^n$$. We then have the zero operators defined with domains \begin{align*} \text{dom}\left(0_{k\times k}\right)&\subseteq H_1\oplus\dots\oplus H_{k},\\ \text{dom}\left(0_{k\times (n-k)}\right)&\subseteq H_1\oplus\dots\oplus H_{k},\\ \text{dom}\left(0_{(n-k)\times k}\right)&\subseteq H_{(k+1)}\oplus\dots\oplus H_{n}, \end{align*} and the block $$M_p$$ defined with domain \begin{align*} \text{dom}\left(M_p\right)&\subseteq H_{(k+1)}\oplus\dots\oplus H_{n}. \end{align*} The operator $$I_{n-k}$$ is the operator $$I_{n-k}= \begin{pmatrix} I_{H_{(k+1)}}&0&\cdots&0\\ 0&I_{H_{(k+2)}}&\cdots&0\\ \vdots&&\ddots&\vdots\\ 0&0&\cdots&I_{H_{n}} \end{pmatrix},$$ where $$I_{H_{i}}$$ is the identity operator on $$H_i$$ for $$i\in\left\{k+1,\dots,n\right\}$$. Does the construction of the restriction of the block operator matrix $$M$$ to the subspace $$V$$ in this case follow analogously with $$M|_V = V^TMV=M_p$$?

There is no difference. One comment, though, is that you removed the original hypothesis that $$M$$ is positive; otherwise, there is a lot of generality lost in assuming that $$M$$ is of the form you say.
When $$M$$ is a positive operator, you can take $$H_0=\ker M$$. Since $$M$$ is selfadjoint, $$\operatorname{ran} M=(\ker M)^\perp$$. So $$H=H_0\oplus H_0^\perp$$. On $$H_0^\perp$$, you have $$\langle Mx,x\rangle >0$$ for all nonzero $$x$$ (otherwise, if $$\langle Mx,x\rangle=0$$, we get $$M^{1/2}x=0$$ and so $$Mx=0$$ so $$x\in H_0$$). That is, $$M$$ is positive-definite on $$H_0^\perp$$. Because $$H_0^\perp$$ is the range of $$M$$, it is trivially invariant for $$M$$; as is $$H_0$$. This implies that $$M$$ is block-diagonal with respect to the decomposition $$H=H_0\oplus H_0^\perp$$. Concretely, if $$P$$ is the projection onto $$\ker M$$, we have $$MP=0$$ (this simply says that if $$Mx=0$$ then $$M^2x=0$$). As $$M=M^*$$, we get $$PM=0$$; thus $$PM=MP=0$$. That is, $$M=(I-P)M(I-P)$$, which is precisely $$M=\begin{bmatrix} 0&0\\0& M_p\end{bmatrix},$$ with $$M_p=(I-P)M(I-P)$$. If you want to isolated $$M_p$$ as an operator, you can do the trick you mention, which is to take $$V:\operatorname{ran}M\to H$$ be the inclusion. It is then easy to see that $$V^*$$ is the orthogonal projection onto $$\operatorname{ran}M$$. Indeed, for any $$x,z\in H$$, $$\tag1 \langle V^*x,Mz\rangle=\langle x,VMz\rangle=\langle x,Mz\rangle.$$ When $$x\in\ker M=(\operatorname{ran}M)^\perp$$, $$(1)$$ tells us that $$V^*x=0$$. When $$x\in\operatorname{ran}M$$, $$(1)$$ tells us that $$x-V^*x=0$$. So $$V^*$$ is indeed the orthogonal projection onto $$\operatorname{ran} M$$. Since $$PV=0$$, $$V=(I-P)V$$. Then $$V^*MV:\operatorname{ran}M\to\operatorname{ran}M$$ satisfies, for any $$x\in\operatorname{ran}M$$, \begin{align} V^*MVx&=V^*(I-P)M(I-P)Vx=V^*(I-P)M(I-P)x\\[0.3cm] &=(I-P)M(I-P)x=M_p. \end{align}
• When you say "$M$ is positive-definite on $H_0$" do you rather mean $M$ is positive-definite on $H_0^{\perp}$, since $H_0$ is the kernel of $M$? Mar 23, 2022 at 10:53